How to proof that nested intervals are an equivalent relation? I want to show that a relation on the space of all sequences of nested intervals is an equivalence relation.

Definition: Let $[a_n,b_n]_{n\in\mathbb{N}}$ and $[c_n,d_n]_{n\in\mathbb{N}}$ be two "sequences of nested intervals".
Define: $[a_n,b_n]_{n\in\mathbb{N}}\simeq [c_n,d_n]_{n\in\mathbb{N}} :\Leftrightarrow \forall n\in\mathbb{N}: a_n\leq d_n \wedge c_n\leq b_n $

with the Definiton for nested Intervalls:

Definition: Let $[a_n,b_n]$ be sequence of Intervalls. We call it nested Intervall $:\Leftrightarrow$a) and b)
a) $\forall n\in\mathbb{N}: a_n\leq a_{n+1}$ and $b_{n+1}\leq b_n $
b) $\{b_n-a_n\}_{n\in\mathbb{N}}$ is a nullsequence

Proof:
Reflexive relation:
Let $[a_n,b_n]_{n\in\mathbb{N}}$ be a sequence of nested intervals.
Because $\forall n\in\mathbb{N}: a_n\leq b_n$ follows  $[a_n,b_n]_{n\in\mathbb{N}}\simeq [a_n,b_n]_{n\in\mathbb{N}}$
Symmetric Relation:
Let  $[a_n,b_n]_{n\in\mathbb{N}}\simeq [c_n,d_n]_{n\in\mathbb{N}}$.
$\Leftrightarrow \forall n\in\mathbb{N}a_n\leq d_n \wedge c_n\leq b_n$
We have to show, that  $[c_n,d_n]_{n\in\mathbb{N}}\simeq [a_n,b_n]_{n\in\mathbb{N}}$, wich meens  $c_n\leq b_n \wedge  a_n\leq d_n$.
This follows immediately.
Transitive relation:
Let  $[a_n,b_n]_{n\in\mathbb{N}}\simeq [c_n,d_n]_{n\in\mathbb{N}}$ and  $[c_n,d_n]_{n\in\mathbb{N}}\simeq [e_n,f_n]_{n\in\mathbb{N}}$
$\Leftrightarrow  \forall n\in\mathbb{N}: a_n\leq d_n \wedge c_n\leq b_n \wedge c_n\leq f_n \wedge e_n\leq d_n $
...?
//edit// I can show that there is at least one $a_n \leq s \leq b_n$ for every 'nested intervall' quite simple... maybe i should use that fact?
...

I just dont see how to mange the transitive part. I think it cant be too hard.
Thanks for your help!

 A: It's true. I'll show transitivity. Define the relation between sequences of closed intervals as:
$$\begin{align}
([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N} \iff & \forall n\in\Bbb N: \\
&\quad a_n \le d_n, \text{ and }\tag{i} \\
&\quad c_n\le b_n.\tag{ii}
\end{align}$$
A sequence of nested intervals is a sequence with the properties (a) and (b). These properties imply that there is a unique point $L$ in the intersection of the intervals, which moreover equals the sup of the left endpoints, the inf of the right endpoints, and the limit of both endpoint sequences:

Any sequence of nested intervals $([a_n,b_n])_{n\in\Bbb N}$ converges to a limit $L$, in several ways:
  $$\begin{align}\tag{$\dagger$}\\
L &= \lim_n a_n = \sup_n a_n\\
&= \lim_n b_n = \inf_n b_n \\
&= \text{the unique element of }\bigcap_n [a_n,b_n].
\end{align}$$

This is fairly well known; proof or reference on request.
Claim: $\simeq$-related sequences of nested intervals converge to the same limit:  

For sequences of nested intervals $([a_n,b_n])_{n\in\Bbb N}$, $([c_n,d_n])_{n\in\Bbb N}$,
  $$if  ([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N}
,\, then 
\, \lim_n a_n = \lim_n c_n = \lim_n b_n = \lim_n d_n.\tag{*}$$

Indeed, given such sequences, let $L=\lim_n a_n$; then we have:
$$\begin{align}
L = \lim_n a_n &\le \lim_n d_n\tag{by (i)} \\
&= \lim_n c_n\tag{by ($\dagger$)} \\
&\le \lim_n b_n\tag{by (ii)} \\
&= L.\tag{by ($\dagger$)} \\
\end{align}$$
Finally, 

Given sequences of nested intervals
  $$([a_n,b_n])_{n\in\Bbb N} \simeq ([c_n,d_n])_{n\in\Bbb N} \simeq ([e_n,f_n])_{n\in\Bbb N},\, 
$$
  we have, for every $m\in\Bbb N$, 
  $$
a_m \le f_m \text{ and } e_m\le b_m\tag{$result$};
$$
  therefore, 
  $$
([a_n,b_n])_{n\in\Bbb N} \simeq ([e_n,f_n])_{n\in\Bbb N}. 
$$

By hypothesis, all six endpoint sequences converge to the same limit $L$, which also equals the $\sup$s and $\inf$s of left-and right-hand endpoint sequences respectively. Given $m$, we have, by (*), $$a_m \le \sup_n a_n = L = \inf_n f_n \le f_m,$$and similarly, $$e_m \le \sup_n e_n = L = \inf_n b_n \le b_m;$$hence the result. But that result, true for all $m$, says exactly that $\simeq$ holds between the sequences $([a_n, b_n])$ and $([e_n, f_n])$, so the conclusion follows.
